Question

Solve the equation explicitly. $$ x y^{3} y^{\prime}=y^{4}+x^{4} $$

Short answer

Answer: $$y(x) = \left(-\frac{3}{4} \ln|y^3 + x^4| - 3C_1\right)^{-\frac{1}{3}}.$$

Step-by-step solution

Rewrite the equation in a separable form

The given equation is: $$ x y^{3} y^{\prime}=y^{4}+x^{4}. $$ To make it separable, divide both sides by \(xy^3\): $$ y^\prime = \frac{y^4 + x^4}{x y^3}. $$ Now the equation is in a separable form.

Separate the variables

Separate the variables by multiplying both sides by \(dx\) and rearrange the terms: $$ \frac{dy}{y^4} = \frac{x^3}{y^3 + x^4} dx. $$

Integrate both sides

Now we can integrate both sides; $$ \int \frac{dy}{y^4} = \int \frac{x^3}{y^3 + x^4} dx. $$ For the left integral, it's straightforward. For the right integral, it seems a bit complicated at first; however, if you notice the differentiation of the denominator, it is related to the numerator. Consider the substitution \(u = y^3 + x^4\). Then, \(du = 4x^3 dx\). So, the right integral becomes \(\frac{1}{4}\int \frac{du}{u}\). Now we can continue to solve the integrals: $$ -\frac{1}{3}y^{-3} = \frac{1}{4} \ln|u| + C_1, $$ where \(C_1\) is an integration constant.

Replace the substitution and solve for y

Replace the substitution \(u = y^3 + x^4\) back into the equation: $$ -\frac{1}{3}y^{-3} = \frac{1}{4} \ln|y^3 + x^4| + C_1. $$ Now, multiply both sides by -3 to isolate \(y^{-3}\): $$ y^{-3} = -\frac{3}{4} \ln|y^3 + x^4| - 3C_1. $$ Raise both sides to the power of -1/3 to find the explicit solution in terms of y: $$ y(x) = \left(-\frac{3}{4} \ln|y^3 + x^4| - 3C_1\right)^{-\frac{1}{3}}. $$ Now, we have found the explicit solution to the given differential equation.

Key concepts

Separation of Variables

Separation of Variables is a technique used to solve differential equations. This method involves rearranging an equation so that each variable and its differential are on separate sides of the equation. For the given equation \( x y^{3} y^{\prime}=y^{4}+x^{4} \), the first step is to express it in a form that makes separating these variables possible.
By dividing both sides by \( xy^3 \), we achieve this: \[ y^\prime = \frac{y^4 + x^4}{x y^3}.\] This setup allows us to manipulate the equation into \[ \frac{dy}{y^4} = \frac{x^3}{y^3 + x^4} dx,\]where each side of the equation involves only one variable and its respective differential. This clears the path for the next steps.

Integration Techniques

Integration is the primary tool used to solve the separated form of a differential equation. In our problem, we need to solve the integrals \[ \int \frac{dy}{y^4} = \int \frac{x^3}{y^3 + x^4} dx.\]For the left integral, the function is a straightforward power of \(y\), that can be solved using the formula for integration of \( y^n \).
- The integral is \( \int y^{-4} dy = -\frac{1}{3} y^{-3} + C_2, \)where \( C_2 \) is a constant.

On the right-hand side, the integration might seem difficult initially. However, by recognizing a pattern between the derivative of the denominator and the numerator, it becomes plausible.
Through substitution, we simplify it, which aids in integrating more complex expressions.

Substitution Method

Sometimes integration can be made easier by substituting a more complex term with a single variable. This is known as the substitution method. In this exercise, substituting helps simplify the right side of the equation. By letting \( u = y^3 + x^4 \), it follows that \( du = 4x^3 dx \).
This reduces our integral to a more manageable form:\[ \frac{1}{4}\int \frac{du}{u},\]which results as:\[ \frac{1}{4} \ln|u| + C_1.\]Here, \( C_1 \) is the constant of integration. This simplifying step is crucial, as it transforms a potentially complicated integral into a simpler logarithmic form.

Explicit Solution

Finding an explicit solution involves expressing the dependent variable solely in terms of the independent variable. After solving the integrals, we substitute back to reclaim any previously introduced variables.
Returning to the original variables, we replace \( u = y^3 + x^4 \) in the integrated equation:\[ -\frac{1}{3}y^{-3} = \frac{1}{4} \ln|y^3 + x^4| + C_1.\]Isolating \( y^{-3} \), and then solving for \( y \) involves algebraic manipulation:\[y(x) = \left(-\frac{3}{4} \ln|y^3 + x^4| - 3C_1\right)^{-\frac{1}{3}}.\] This gives us a clear expression for \( y \) as a function of \( x \), providing a complete solution to the differential equation. This method ensures clarity and completeness in our solution.